It was shown in the answer of the problem what is the right way to calclulate $\partial_{z}\frac{1}{\bar{z}?}$ that \begin{align} \partial_{z}\frac{1}{\bar{z}}=\pi\delta^{(2)}(\vec{x}), \end{align} but when I tried to calclulate $\partial_{z}\frac{1}{\bar{z}^2}$ further, I found that the answer is depend on the method I used.
Method $1:$ \begin{align} \partial_{z}\frac{1}{\bar{z}^2}&=\partial_{z}(\frac{1}{\bar{z}}\cdot\frac{1}{\bar{z}})\\ &=2\cdot\frac{1}{\bar{z}}\partial_{z}\frac{1}{\bar{z}}\\ &=\frac{2\pi}{\bar{z}}\delta^{(2)}(\vec{x}). \end{align} Method $2:$ \begin{align} \partial_{z}\frac{1}{\bar{z}^2}&=-\partial_{z}\partial_{\bar{z}}\frac{1}{\bar{z}}\\ &=-\partial_{\bar{z}}\partial_{z}\frac{1}{\bar{z}}\\ &=-\pi\partial_{\bar{z}}\delta^{(2)}(\vec{x}). \end{align} I tried to prove that the two answers were equivalent because both methods seem natural, and my starting point is $x\delta'(x)=-\delta(x)$, where prime means derivative with respect to x, but it failed.
comment : It seems physicist prefer the second answer (I'm a physics student and my tutor provides me with method 2, but he can't explain why method 1 is wrong.)
Thank you for any help.